Q143 / 160 · TS EAMCET 2019 (06 May Shift 1) · Physics

An object $O$ is placed at $7 \mathrm{~cm}$ to the left of a concave mirror of radius of curvature $12 \mathrm{~cm}$ as shown in the figure. The position of the image will be at a distance of Figure

  • A
    $20 \mathrm{~cm}$ from the mirror to the left.
  • B
    $30 \mathrm{~cm}$ from the mirror to the right.
  • Correct answer
    C
    $42 \mathrm{~cm}$ from the mirror to the left.
  • D
    $42 \mathrm{~cm}$ from the mirror to the right.

Solution

Given, distance between object and mirror, $u=-7 \mathrm{~cm}$ Radius of curvature, $R=-12 \mathrm{~cm}$ $\therefore$ Focal length, $f=-\frac{12}{2}=-6 \mathrm{~cm}$ From mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ Putting the given values, $\begin{array}{cc} & \frac{1}{v}+\frac{1}{-7}=\frac{1}{-6} \\ & \frac{1}{v}=\frac{1}{-6}+\frac{1}{7}=-\left(\frac{1}{6}-\frac{1}{7}\right) \\ \Rightarrow \quad v & v=42 \mathrm{~cm}\end{array}$ Hence, the position of the image will be $42 \mathrm{~cm}$ from the mirror to the left.

Hence, the correct answer is option: 'C'.

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